One end of a light string is fixed to a clamp | vedclass

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(C) Let $T$ be the tension in the string.The maximum vertical force on the clamp is given by $T \sin 30^{\circ} = 40 \,N$.Substituting the value of $\

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$6$

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(C) Let $T$ be the tension in the string.The maximum vertical force on the clamp is given by $T \sin 30^{\circ} = 40 \,N$.Substituting the value of $\sin 30^{\circ} = 1/2$:$T \cdot (1/2) = 40$$T = 80 \,N$Now,consider the free-body diagram of the monkey of mass $m = 5 \,kg$ climbing with acceleration $a$.The forces acting on the monkey are tension $T$ upwards and weight $mg$ downwards.According to Newton's second law: $T - mg = ma$$a = (T - mg) / m$Substituting the values $T = 80 \,N$,$m = 5 \,kg$,and $g = 10 \,ms^{-2}$:$a = (80 - 5 	imes 10) / 5$$a = (80 - 50) / 5$$a = 30 / 5 = 6 \,ms^{-2}$Thus,the maximum acceleration is $6 \,ms^{-2}$.

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